![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
tính riêng:
\(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\)
=\(\left(\frac{100}{99}-1\right)+\left(\frac{100}{98}-1\right)+\left(\frac{100}{97}-1\right)+...+\left(\frac{100}{2}-1\right)+99\)
=\(100.\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)+99-98\)
=\(100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)\)
vậy \(\left(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=100\)
chúc bạn học tốt ^^
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
=>\(3B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
=>\(3B-B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-...-\dfrac{1}{3^{99}}\)
=>\(2B=1-\dfrac{1}{3^{99}}\)
=>\(2B=\dfrac{3^{99}-1}{3^{99}}\)
=>\(B=\dfrac{3^{99}-1}{3^{99}\cdot2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(=\frac{\left(101+1\right).100:2}{\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1}\)
\(=\frac{5050}{1+1+...+1+1}\)(51 chữ số 1)
= \(\frac{5050}{51}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012
2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013
3M=2^0+2^2013
M=(2^0+2^2013)÷3
Vậy.......
b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012
3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013
4N=3-3^2013
N=(3-3^2013)÷4
Vậy........
K tao nhé ko lên lớp tao đánh m😈😈😈
![](https://rs.olm.vn/images/avt/0.png?1311)
A=3^100-3^99+3^98-3^97+.................+3^2-3+1
3A = 3^101-3^100+3^99-3^98+3^97-3^96+...........................-3^2+3
3A + A = 3101+1
4A = 3101 + 1
A = \(\frac{3^{101}+1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = 2 ^ 100 + 2 ^ 99 + 2 ^ 98 + ... + 2 ^ 2 + 2 ^ 1
2A = 2 ^ 101 + 2 ^ 100 + 2 ^ 99 + ... + 2 ^ 3 + 2 ^ 2
2A - A = ( 2 ^ 101 + 2 ^ 100 + 2 ^ 99 + ... + 2 ^ 3 + 2 ^ 2 )
- ( 2 ^ 100 + 2 ^ 99 + 2 ^ 98 + ... + 2 ^ 2 + 2 ^ 1 )
A = 2 ^ 101 - 2
\(A=2^{100}+2^{99}+2^{98^{ }}+...+2^2+2^1\)
\(2A=2.\left(2^{100}+2^{99}+...+2^1\right)\)
\(2A=2^{101}+2^{100}+...+2^2+2^1\)
\(A=2A-A\)
\(A=2^{101}-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xin lỗi, nhìn nhầm:
A = 3^100 - 3^99 + 3^98 - 3^97 +...........+ 3^2 - 3 + 1
3A = 3^101 - 3^100 + 3^99 - 3^98 +...+3^3 -3^2 +3
=> 4A = 3A + A = 3^101 + 1
A = \(\frac{3^{101}+1}{4}\)
B = 3^100 - 3^99 + 3^98 - 3^97 +...........+ 3^2 - 3 + 1
3B = 3^101 - 3^100 + 3^99 - 3^98 +...+3^3 -3^2 +3
Cộng vế với vế triệt tiêu, ta có :
4B = 3^101 + 1
B = \(\frac{3^{101}+1}{4}\)
RÚT GỌN LÀ RÚT GỌN THẾ NÀO HẢ BẠN
99 + 277= 318+321 = 318(1+33)
96 + 2433= 312+315 = 312 (1+33)